In the right triangle shown, $\angle B = 60^\circ$ and $AC = 6\sqrt{3}$. How long is $AB$ ? $A$ $C$ $B$ $6\sqrt{3}$ $x$
We know the length of a leg, and want to find the length of the hypotenuse. What mathematical relationship is there between a right triangle's legs and its hypotenuse? We can use either sine (opposite leg over hypotenuse) or cosine (adjacent leg over hypotenuse). This is a 30-60-90 triangle, so we know what the values of sine and cosine are at each angle of the triangle. Let's try using cosine: $A$ $C$ $B$ $6\sqrt{3}$ $x$ ${30}^{\circ}$ Cosine is adjacent over hypotenuse (SOH CAH TOA), so $\cos {30}^{\circ} = \dfrac{6\sqrt{3}}{x}$ . We also know that $\cos{30}^{\circ} = \dfrac{\sqrt{3}}{2}$ Solving for $x$ , we get $ x \cdot \cos{30}^{\circ} = 6\sqrt{3}$ $ x \cdot \dfrac{\sqrt{3}}{2} = 6\sqrt{3}$ $ x = 6\sqrt{3} \cdot \dfrac{2}{\sqrt{3}}$ $ x = 6\sqrt{3} \cdot \dfrac{2\cdot\sqrt{3}}{3}$ So, $x = 12$.